Chapter 3 Genetics Punnet Sqaure/Grid Practice
Multiple Choice
Identify the choice that best completes the statement or answers the question.
RrYy
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
RrYy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Seed Shape
R = Round
r = Wrinkled
Seed Color
Y = Yellow
y = Green
Figure 11–2
____
1. The Punnett square in Figure 11–2 shows that the gene for pea shape and the gene for pea color
a. assort independently.
b. are linked.
c. have the same alleles.
d. are always homozygous.
Figure 14–1
____
2. According to Figure 14–1, what is the approximate probability that a human offspring will be female?
a. 10%
b. 25%
c. 50%
d. 75%
Blood Type
A
B
AB
O
Blood Types
Combination of Alleles
IAIA or IAi
IBIB or IBi
IAIB
ii
Figure 14–2
____
3. If a man with blood type A and a woman with blood type B produce an offspring, what might be the
offspring’s blood type?
a. AB or O
b. A, B, or O
c. A, B, AB, or O
d. AB only
Figure 14–3
____
4. Examine the pedigree in Figure 14–3. The allele for the presence of a white forelock is dominant. What is the
probability of the couple labeled 2 of having a child with a white forelock?
a. 25%
b. 50%
c. 75%
d. 100%
Completion
Complete each statement.
1. A(An) ____________________ is a diagram that follows the inheritance of a single gene through several
generations of a family.
Short Answer
RrYy
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
RrYy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Seed Shape
R = Round
r = Wrinkled
Seed Color
Y = Yellow
y = Green
Figure 11–2
1. What is the phenotype ratio of the offspring of the plants in the Punnett square in Figure 11–2?
2. A tall pea plant with yellow seeds is heterozygous for height and seed color (TtYy). This plant is crossed
with a pea plant heterozygous for height but homozygous recessive for seed color (Ttyy). If 80 offspring are
produced, how many are expected to be tall and have yellow seeds?
Science Skills
The pedigree shows the inheritance of free earlobes and attached earlobes in five generations of a family.
Attached earlobes are caused by a recessive allele (f).
Figure 14–13
1. Infer Is individual 2 in Figure 14–13 homozygous or heterozygous for free earlobes? Explain.
2. Interpret Visuals In Figure 14–13, how many children of individuals 4 and 5 have attached earlobes?
3. Infer Can you be certain of the genotype of individual 5 in Figure 14–13? Explain.
4. Predict Predict the genotype and phenotype of individual 14 in Figure 14–13.
5. Infer In Figure 14–13, are any of the descendants of individuals 1 and 2 homozygous for free earlobes?
Chapter 3 Genetics Punnet Sqaure/Grid Practice
Answer Section
MULTIPLE CHOICE
1. ANS:
OBJ:
TOP:
2. ANS:
OBJ:
STA:
3. ANS:
OBJ:
STA:
4. ANS:
OBJ:
STA:
BLM:
A
PTS: 1
DIF: L2
REF:
11.2.2 Explain the principle of independent assortment.
STA:
Foundation Edition
BLM: comprehension
C
PTS: 1
DIF: L1
REF:
14.1.1 Identify the types of human chromosomes in a karyotype.
CA.BIO.2.f | CA.BIO.2.g
BLM: comprehension
C
PTS: 1
DIF: L3
REF:
14.1.2 Describe the patterns of the inheritance of human traits.
CA.BIO.3.a BLM: analysis
C
PTS: 1
DIF: L3
REF:
14.1.3 Explain how pedigrees are used to study human traits.
CA.BIO.3.c*
TOP: Foundation Edition
evaluation
p. 317
CA.BIO.3.a | CA.BIO.3.b
p. 392 | p. 393
p. 394
p. 396 | p. 397
COMPLETION
1. ANS: pedigree
PTS: 1
DIF: L1
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
BLM: knowledge
SHORT ANSWER
1. ANS:
The phenotype ratio is 9 round, yellow seeds : 3 round, green seeds : 3 wrinkled, yellow seeds : 1 wrinkled,
green seed.
PTS: 1
DIF: L2
REF: p. 317
OBJ: 11.2.2 Explain the principle of independent assortment.
STA: CA.BIO.3.a | CA.BIO.3.b
TOP: Foundation Edition
BLM: application
2. ANS:
Thirty of the offspring are expected to be tall and have yellow seeds.
PTS: 1
DIF: L3
REF: p. 316 | p. 317
OBJ: 11.2.2 Explain the principle of independent assortment.
STA: CA.BIO.3.a | CA.BIO.3.b
BLM: analysis
SCIENCE SKILLS
1. ANS:
The individual is heterozygous (Ff), since her daughter has attached earlobes. The daughter inherited one
allele for attached earlobes from individual 2 and another from individual 1.
PTS: 1
DIF: L2
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
TOP: Foundation Edition
BLM: analysis
2. ANS:
None of those children have attached earlobes.
PTS: 1
DIF: L2
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
TOP: Foundation Edition
BLM: application
3. ANS:
No, the genotype of individual 5 is uncertain because his children have free earlobes. Thus, individual 5 could
be homozygous (FF) or heterozygous (Ff) for free earlobes. If, however, one of his children had attached
earlobes, it would be certain that individual 5 was heterozygous.
PTS: 1
DIF: L3
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
BLM: analysis
4. ANS:
Individual 14 will be homozygous for the recessive allele (ff) and will have attached earlobes.
PTS: 1
DIF: L2
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
BLM: analysis
5. ANS:
No, none of the descendants are homozygous (FF) for free earlobes.
PTS: 1
DIF: L3
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
BLM: evaluation